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On the Number of Integers Representable as Sums of Unit Fractions, III
Journal of Number Theory 96, 351–372 (2002) doi:10.1006/jnth.2002.2797
On the Number of Integers Representable as Sums of Unit Fractions, III Hisashi Yokota Department of Mathematics, Hiroshima Institute of Technology, 2-1-1 Miyake Saeki-ku, Hiroshima, Japan E-mail: [email protected] Communicated by D. Goss Received July 16, 2001; revised January 4, 2002
Let NðnÞ be the set of all integers that can be expressed as a sum of reciprocals of distinct integers 4n: Then we prove that for sufficiently large n; log n þ g
2 p ðlog2 nÞ2 þ oð1Þ 4jNðnÞj; 3 log n
which improves the lower bound given by Croot.
1.
# 2002 Elsevier Science (USA)
INTRODUCTION
Let NðnÞ be the set of all integers a that can be expressed as a¼
X ek ; k 14k4n
where ek takes either 0 or 1: Answering questions of Erdo+ s and Graham [5], we have shown (see [10–13]) that 1 ðlog2 nÞ2 þ oð1Þ : log n þ g 2 oð1Þ4jNðnÞj4log n þ g 4 log n Then recently, in [2], Croot improved our previous results so that
9 ðlog2 nÞ2 1 ðlog2 nÞ2 þ oð1Þ þ oð1Þ log n þ g 4jNðnÞj4log n þ g : 2 2 log n log n Here and in the sequel, we let logj denote the jth fold iterated logarithm. 351 0022-314X/02 $35.00 # 2002 Elsevier Science (USA) All rights reserved.
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HISASHI YOKOTA
In connection with NðnÞ; Don Zagier, in private communication, pointed out the following: Let a be a given integer and define F ðaÞ ¼ minfn : a 2 NðnÞg: Then determining NðnÞ for all n is the same as calculating F ðaÞ since a 2 NðnÞ iff n5F ðaÞ: He then asked if it is possible to improve the upper bound of F ðaÞ: Let S :¼ fsj : j51g be the increasing sequence of all positive integers of i the form p2 ; i50; where p a prime. Once st is chosen, we denote pkðtÞ the largest prime os2t ; puðtÞ the smallest prime > st : We define 8 9 t kðtÞ < = Y Y DðtÞ ¼ d4LðtÞ: d si pj ; : ; 1 uðtÞ where LðtÞ ¼ pkðtÞ ðlog pkðtÞ Þ2 ðlog2 pkðtÞ Þ2 : Now we ask how many integers a can be expressed as a¼
X
ek ; k k2DðtÞ;k4n
where ek takes either 0 or 1: Let N n ðnÞ be the set of all integers a that can be expressed as a¼
X k2DðtÞ;k4n
ek ; k
where ek takes either 0 or 1 and F n ðaÞ ¼ minfn: a 2 N n ðnÞg: Then N n ðnÞ NðnÞ and jN n ðnÞj4jNðnÞj: In this paper, we prove the following results. Theorem 1.
There exists a constant n0 such that, for all n > n0 ;
p2 ðlog2 nÞ2 þ oð1Þ 4jN n ðnÞj log n þ g 3 log n and "
# p2 ðlog aÞ2 F ðaÞ4exp a g þ þ oð1Þ : 3 a n
By noting jN n ðnÞj4jNðnÞj and F ðaÞ4F n ðaÞ; we improve the lower bound of jNðnÞj and the upper bound of F ðaÞ: Also by quoting Croot’s result about
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
353
the upper bound of jNðnÞj; we obtain Corollary 1. There exists a constant n0 such that, for all n > n0 ; p2 ðlog2 nÞ2 1 ðlog2 nÞ2 þ oð1Þ þ oð1Þ 4jNðnÞj4log n þ g log n þ g 2 3 log n log n
and "
# 2 p ðlog aÞ2 þ oð1Þ F ðaÞ4F ðaÞ4exp a g þ : 3 a n
2. Lemma 1.
LEMMATA 2=5
For all large pkðtÞ ; pkðt1Þ 5pkðtÞ ð1 21=5 Þ: pkðtÞ
Lemma 2. There exists a constant t0 such that, for all t5t0 and ðlog2 pk Þ2 Qt all r satisfying r ¼ ð1 þ oð1ÞÞ there are distinct 1 si ; Q P log pk Q integers di with the property that r ¼ di ; where di j t1 si and di 5 t1 si =pk log pk ð1 log2 pk Þ: 2
Lemma 3. There exists a constant2 Q t0 such Qk that, for all t5t0 and t 2 pk Þ all r satisfying r ¼ ð1 þ oð1ÞÞðlog s 1 i u pj ; QthereQ are distinct log pP k integers d with the property that r ¼ d ; where di j t1 si ku pj and di 5 i Qt Qk i 1 si u pj =LðtÞ: Lemma 4.
There exists a constant t0 such that for all t5t0
2 X ðlog2 pkðtÞ Þ2 ðlog2 pkðtÞ Þ2 15 p2 1 p 3 4 þ oð1Þ þ oð1Þ 4 : d 3 3 log pkðtÞ log pkðtÞ d4LðtÞ;deDðtÞ
Lemma 5.
There exists a constant t0 such that for all t5t0 X 1 X 1 log2 pkðtÞ 4ð4 þ oð1ÞÞ 2 : d d log pkðtÞ d4LðtÞ d4Lðt1Þ d2DðtÞ
d2Dðt1Þ
Assuming these lemmas hold, we prove Theorem 1.
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HISASHI YOKOTA
3.
PROOF OF THEOREM 1
Given a large integer a; we choose t such that X d4Lðt1Þ d2Dðt1Þ
X 1 1 ½ðlog aÞ2 oa þ o : d a d d4LðtÞ d2DðtÞ
Since by Lemmas 4 and 5,
aþ
LðtÞ X 1 X X 1 ½ðlog aÞ2 1 o ¼ a d j d4LðtÞ d j¼1 d4LðtÞ d2DðtÞ
deDðtÞ
ðlog2 pkðtÞ Þ2 15 p2 þ oð1Þ 4 log LðtÞ þ g 3 log pkðtÞ and
aþ
X 1 X 1 log2 pkðtÞ ½ðlog aÞ2 > 5 ð4 þ oð1ÞÞ 2 a d d4LðtÞ d log pkðtÞ d4Lðt1Þ d2Dðt1Þ
d2DðtÞ LðtÞ X 1 X log2 pkðtÞ 1 ð4 þ oð1ÞÞ 2 ¼ j d log pkðtÞ j¼1 d4LðtÞ deDðtÞ
5
2 LðtÞ X ðlog2 pkðtÞ Þ2 1 p 3 þ oð1Þ j 3 log pkðtÞ j¼1
2 ðlog2 pkðtÞ Þ2 p 3 þ oð1Þ ¼ log LðtÞ þ g : 3 log pkðtÞ 2
Let a0 ¼ log LðtÞ þ g ðp3 þ oð1ÞÞ ðlog2 pkðtÞ Þ2 ; we have
ðlog2 pkðtÞ Þ2 log pkðtÞ :
Then since LðtÞ ¼ pkðtÞ ðlog pkðtÞ Þ2
2 ðlog2 pkðtÞ Þ2 ½ðlog a0 Þ2 p 3 þ oð1Þ a0 þ ¼ log LðtÞ þ g : a0 3 log pkðtÞ
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
355
2 ðlog2 pkðtÞ Þ2 p þ oð1Þ a5a0 ¼ log LðtÞ þ g : 3 log pkðtÞ
ð1Þ
Thus,
On the we have
other
hand,
for
2
b0 ¼ log LðtÞ þ g ð18p 3 þ oð1ÞÞ
ðlog2 pkðtÞ Þ2 log pkðtÞ ;
ðlog2 pkðtÞ Þ2 ½ðlog b0 Þ2 15 p2 b0 þ þ oð1Þ ¼ log LðtÞ þ g : b0 3 log pkðtÞ Thus, a4b0 ¼ log LðtÞ þ g
ðlog2 pkðtÞ Þ2 18 p2 þ oð1Þ : 3 log pkðtÞ
This shows that a ¼ ð1 þ oð1ÞÞ log pkðtÞ : Now we show that X ej a¼ : j j4LðtÞ j2DðtÞ
Since a ¼ ð1 þ oð1ÞÞ log pkðtÞ ; a 2 DðtÞ; and by Lemma 5, we have X 1 d d4LðtÞ
! log2 pkðtÞ ½ðlog aÞ2 r aþ : ¼ Qt QkðtÞ 4ð4 þ oð1ÞÞ 2 a log pkðtÞ 1 si uðtÞ pj
d2DðtÞ
Rewriting this, we have 0 1 2 X 1 r ½ðlog aÞ
A @Qt QkðtÞ þ a¼ d a s p d4LðtÞ 1 i uðtÞ j d2DðtÞ
0 1 n X 1 r ¼ @Qt QkðtÞ A; d d4LðtÞ 1 si uðtÞ pj d2DðtÞ
where rn satisfies the following: rn ¼ ð1 þ oð1ÞÞ
kðtÞ t Y ðlog2 pkðtÞ Þ2 Y si pj : log pkðtÞ 1 uðtÞ
ð2Þ
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HISASHI YOKOTA
Then by Lemma 3, we can express rn as a sum of distinct divisors Qt QkðtÞ si pj Q P QkðtÞ di of t1 si uðtÞ pj with di 5 1 LðtÞ uðtÞ such that rn ¼ m 1 di : Thus, Pm X 1 X ej 1 di Qt Q a¼ : ¼ kðtÞ d j d4LðtÞ j4LðtÞ 1 si uðtÞ pj d2DðtÞ
j2DðtÞ
This shows that for every large integer a; we can find t so that a is a sum of reciprocals of distinct integers in DðtÞ less than or equal to LðtÞ; where LðtÞ satisfies 2 ðlog2 pkðtÞ Þ2 p þ oð1Þ a5log LðtÞ þ g 3 log pkðtÞ and ðlog2 pkðtÞ Þ2 18 p2 þ oð1Þ : a4log LðtÞ þ g 3 log pkðtÞ We note that a¼
X ej j j4LðtÞ j2DðtÞ
implies that a 2 N n ðLðtÞÞ ð1 þ oð1ÞÞ log pkðtÞ ; we have
and
F n ðaÞ4LðtÞ:
Then
since
a¼
2 ðlog2 pkðtÞ Þ2 p2 p ðlog aÞ2 þ oð1Þ þ oð1Þ : a5log LðtÞ þ g ¼ log LðtÞ þ g 3 log pkðtÞ 3 a
Thus, "
# p2 ðlog aÞ2 F ðaÞ4LðtÞ4exp a g þ þ oð1Þ : 3 a n
We also note that a 2 N n ðLðtÞÞ implies
ðlog2 pkðtÞ Þ2 p2 þ oð1Þ log LðtÞ þ g 4jN n ðLðtÞÞj: 3 log pkðtÞ
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
357
Let n be an integer such that LðtÞ4noLðt þ 1Þ: Then log LðtÞ4log n olog Lðt þ 1Þ and by Lemma 1 log Lðt þ 1Þ ¼ ð1 þ oð1ÞÞ log pkðtþ1Þ 0
0
11 1 ¼ ð1 þ oð1ÞÞ@log pkðtÞ þ O@ 1=5 AA pkðtÞ 0 1 1 ¼ log LðtÞ þ O@ 1=5 A: pkðtÞ
Therefore, log n þ g
4.
2 p ðlog2 nÞ2 þ oð1Þ 4jN n ðnÞj: 3 log n
PROOF OF LEMMAS 3=5
Since for pi large, by [4], we have piþ1 4pi þ pi :
Proof of Lemma 1.
3=5
6=5
Thus pkðt1Þ 5pkðt1Þþ1 pkðt1Þ 5s2t1 st1 : By the choice of st ; we have 3=5
st1 5st st1 : This then implies that pkðt1Þ 5ðst
¼ s2t 1
3=5 st1 Þ2
22=5
!
4=5
st
6=5 st 5s2t
0 5pkðtÞ @1
8=5 2st 5s2t
1 22=5 A 1=5
:
2=5 ! 2 1 st
]
pkðtÞ
l
Proof of Lemma 2. Case 1: st ¼ p2 ; l51: Define for j ¼ 1; 2; 3; . . . ; 2l ; (Q ) t1 Y t1 si j j 1 : p p þ 14d4p 1; d Dj ¼ s [ f0g: 1 i p2l j d We note that if dj 2 Dj and dj a0; then for all i; Qt1
Qt1 si si 4dj 4 2l j j1 : l j 2 j p ðp 1Þ p ðp p þ 1Þ 1
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HISASHI YOKOTA
Q Q Q 2l 1 Then by the property of S; t1 ti ; where p[ ti for all i: 1 si ¼ p Thus, (Q ) t1 Y t1 si j j 1 : p p þ 14d4p 1; d s [ f0g Dj ¼ 1 i p2l j d ( ¼ pj1
t1 ) Y : pj p þ 14d4pj 1; d si [ f0g d 1
Q
ti
f0; pj1 ; 2pj1 ; . . . ; ðp 1Þpj1 g ðmod pj Þ: Pl Let Dn ¼ 21 Dj : Then by Lorentz’s theorem [6], Dn is a complete n residue system P l modulo st : Thus given any integer r; r dt ðmod st Þ for some dtn ¼ 21 dj ej 2 Dn ; where dj 2 Dj ; and dtn 4
t1 Y
si
2l X
1
1
1 p2l j ðpj p þ 1Þ
!
t1 Y 2 log st si 4 : st log 2 1 Case 2: st ¼ p: Define (Q D¼
) Y t1 si st 1 4dost ; d : s [ f0g: 1 i 2 d
t1 1
Then by Lemma 1 of [9], jDj ¼ pþ3 2 and no two elements are congruent modulo st : If r 2d ðmod st Þ for some d 2 D; then let Dn ¼ ðD fdgÞ þ ðD fdgÞ: Otherwise, Dn ¼ D þ D:PThen by Cauchy–Davenport Theorem [3], r dtn ðmod st Þ for some dtn ¼ 21 dj ej 2 Dn ; where dj 2 D; and n
dt 4
t1 Y 1
4
t1 Y 1
2 2 þ si st 1 st þ 1
2 log st si : st log 2
Thus in either case, given any integer r; we have r dtn ðmod st Þ and dtn 4
t1 Y 1
si
2 log st : st
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
359
Now given r such that r ¼ ð1 þ oð1ÞÞ
t ðlog2 pkðtÞ Þ2 Y si : log pkðtÞ 1
We let r1 ¼ Then
r dtn ; st
an integer:
! t1 ðlog2 pkðtÞ Þ2 2 log st Y r1 5 ð1 þ oð1ÞÞ 2 si log pkðtÞ st log 2 1
and r1 4ð1 þ oð1ÞÞ
t1 ðlog2 pkðtÞ Þ2 Y si : log pkðtÞ 1
Continue this process j times so that stj is the smallest element in S satisfying log pkðtÞ ð1 log 2pkðtÞ Þ4stj : Then, 2 ! tj t ðlog2 pkðtÞ Þ2 2 X log si Y rj 5 ð1 þ oð1ÞÞ si : 2 log 2 tj si log pkðtÞ 1 Since Z 2pffiffiffiffiffiffi t pkðtÞ X log si 4 s2i log pkðtÞ ð1 tj
2 log2 pkðtÞ
2pffiffiffiffiffiffi pkðtÞ
log x 1 ¼ x x ¼ ð1 þ oð1ÞÞ
log2 pkðtÞ log x dx þ O 2 ðlog pkðtÞ Þ2 Þ x þO
log pkðtÞ ð12=log2 pkðtÞ Þ
!
log2 pkðtÞ
!
ðlog pkðtÞ Þ2
log2 pkðtÞ ; log pkðtÞ
! tj ðlog2 pkðtÞ Þ2 log2 pkðtÞ Y 2 ð1 þ oð1ÞÞ rj 5 ð1 þ oð1ÞÞ si log 2 log pkðtÞ log pkðtÞ 1 ! tj ðlog2 pkðtÞ Þ2 Y ¼ ð1 þ oð1ÞÞ si : log pkðtÞ 1
360
HISASHI YOKOTA
We also note that tj ðlog2 pkðtÞ Þ2 Y rj 4ð1 þ oð1ÞÞ si : log pkðtÞ 1
Thus, rj ¼ ð1 þ oð1ÞÞ
tj ðlog2 pkðtÞ Þ2 Y si : log pkðtÞ 1
Q Since by Lemma 1 of [8], every integer a4 tj si can be expressed as a sum 1 Qtj of distinct divisors of 1 si : Thus, we can express rj as follows: tj Y X rj ¼ di0 ; where di0 si ; di0 51: 1
Then r ¼ st r1 þ dtn n ¼ st ðst1 r2 þ dt1 Þ þ dtn
.. . ¼
!
t Y
si rj þ
tjþ1
¼
t Y tjþ1
si
!
t Y
n si dtjþ1 þ
tjþ2
X
di0 þ
!
t Y
n si dtjþ2 þ þ dtn
tjþ3 t Y
! n si dtjþ1 þ
tjþ2
t Y
! n si dtjþ2 þ þ dtn
tjþ3
Q Q n and ttmþ1 si dtm are allPdistinct for m ¼ 0; 1; 2; . . . ; j: Here ttþ1 si ¼ 1: s n Note that for dtm ¼ 1tm dj ej a0; we have by Cases 1 and 2, Qtm1 1 dj 5stm 1 si : Thus, t t tm1 Y Y 1 Y s i dj > si si stm1 tmþ1 tmþ1 1 Qt 5
Qt si 1 si 5 1 : 2 st 4pkðtÞ
Q Since stj1 4log pkðtÞ ð1 log 2pkðtÞ Þ; we have tj1 si 4pkðtÞ by Lemma 2 of 1 2 [1]. Thus, Qt Qt Qt t Y 1 si 1 si 1 si : si ¼ Qtj ¼ Qtj1 5 si si stj pkðtÞ log pkðtÞ ð1 log22pkðtÞ Þ tjþ1 1 1
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
361
P Q This Qshows that r ¼ di ; where di are distinct divisors of t1 si and di 5 t1 si =pkðtÞ log pkðtÞ ð1 log 2pkðtÞ Þ: ] 2
For u ¼ uðtÞ4j4k ¼ kðtÞ; let
Proof of Lemma 3. ( Dj ¼
) Y t d: 14d4log pj log2 pj ðlog3 pj Þ ; d s : 1 i 5
Let q be the smallest element in S satisfying q > log pkðtÞ : Then qo2 log pkðtÞ : Now define for j ¼ u; u þ 1; . . . ; k (Q
n
Dj ¼ where
Qu1 u
t 1
Qj1
si
u
pi
qd
) : d 2 Dj ;
pi ¼ 1: Note that if djn 2 Dnj ; then djn is an integer and Qt
1
si
Qj1 u
pi
n
q log pj log2 pj ðlog3 pj Þ
5
Qt
4dj 4
1
si
Qj1 u
pi
q
:
We claim that (
mj X
) djn ej : ej ¼ 0 or 1
f1; 2; . . . ; pj g ðmod pj Þ;
1
where mj 4q log pj ðlog2 pj Þ2 : Let a be a residue modulo pj : Choose tn such that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi log pj ð1 2=log2 pj Þ4stn 42 log pj ð1 2=log2 pj Þ: Then by Lemma 1 of [8], tn Y 1
Q tn 1
si
si opj : Thus, we can choose kn so that
n kY 1
un
pi opj o
tn Y 1
si
kn Y
pi ;
un
where pun is the smallest prime greater than stn : Then by Lemma 2 of [1], pffiffiffiffiffiffiffi log pj ð1 2=log2 pj Þ4pkn 4log pj ð1 þ 2=log2 pj Þ: Thus stn =24 pkn 42stn : Now, consider Qn Q n pj s þ rn a a t si kn pi ¼ Q1tn Qukn ¼ Q tn Q k n ; pj pj 1 si un pi pj 1 si un pi
362
HISASHI YOKOTA
where rn is chosen so that t k Y pffiffiffiffiffi Y ð1 þ 1=log2 pkn 2= stn Þ si pi n
n
1
un
t k Y pffiffiffiffiffi Y 4rn oð2 þ 1=log2 pkn 2= stn Þ si pi : n
n
1
un
P
n Then by [9], di ; where di are distinct divisors of QknTheorem 2.2Qof Qkrn ¼ Q tn tn 5 n log pk n ðlog pk n Þ : Thus, s p and d 5 s p =p n n i i i i i k 2 1 u 1 u
a
tn Y
si
1
kn Y
pi r n
un
tn Y
si
kn Y
1
un
tn Y
kn Y
si
1
!
P p i Q tn 1
pi
un
si
mj X ei i i¼1
di Qk n un
pi
! ðmod pj Þ;
Q tn Q k n where mj is less than or equal to pkn log pkn ðlog2 pkn Þ5 : Since 1 nsi un pi ; Q tn ð Q pj Þ ¼ 1; as a runs through all residues modulo pj ; a 1 si kun pi runs through all residues modulo pj : Thus, n ) ( n ! mj Y t kn k Y Y X si pi ei =i : ei ¼ 0 or 1; i p un i 1 i¼1 un contains all residues modulo pj : Furthermore, t Y 1
si
j1 Y u
pi =q
tn Y 1
si
kn Y
! pi ; pj
¼ 1:
un
Thus, we have mj X
djn ej f1; 2; . . . ; pj g ðmod pj Þ;
1
where mj 4qpkn log pkn ðlog2 pkn Þ5 4q log pj ðlog2 pj Þ2 for pj large. Now given r such that ! t k Y ðlog2 pk Þ2 Y r ¼ ð1 þ oð1ÞÞ si pj ; log pk u 1
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
363
we can express r as r
X
dkn ek ðmod pk Þ;
where dkn 2 Dnk
and X
Let r1 ¼ ðr
! mk t k 1 t k 1 Y X Y 1Y ei log mk Y dk e k ¼ si pi si pi : 4 q 1 i q u u i¼1 1 n
P
dkn ek Þ=pk : Then,
r1 5 ð1 þ oð1ÞÞ
t k 1 t k 1 Y Y ðlog2 pk Þ2 Y log mk Y si pi si pi log pk qpk 1 u u 1
ðlog2 pk Þ2 log mk ¼ ð1 þ oð1ÞÞ log pk qpk
!
t Y
si
k 1 Y
1
pi
u
and r1 4
t k 1 Y r ðlog2 pk Þ2 Y ¼ ð1 þ oð1ÞÞ si pi : pk log pk u 1
Repeating the same argument k u times, we have k ðlog2 pk Þ2 log mk X 1 rku 5ð1 þ oð1ÞÞ p log pk q j¼u j
!
t Y
si :
1
2 Since mk oq Pklog 1pk ðlog pk Þ ; we have log mk 4ð2 þ oð1ÞÞ log2 pk : We note that since j¼u p ¼ Oð1Þ;
k log mk X 1 log2 pk ¼O : p q log pk j¼u j Thus, we have ! Y t ðlog2 pk Þ2 log2 pk O si rku 5 ð1 þ oð1ÞÞ log pk log pk 1 ¼ ð1 þ oð1ÞÞ
t ðlog2 pk Þ2 Y si : log pk 1
364
HISASHI YOKOTA
Therefore, rku ¼ ð1 þ oð1ÞÞ
t ðlog2 pk Þ2 Y si : log pk 1
Note that X r ¼ pk r 1 þ dkn ek X X n ¼ pk pk1 r2 þ dk1 ek þ dkn ek ¼
k Y u
and
Qk j
pi rku þ
k Y
pi
X
X X n dun eu þ þ pk dk1 ek1 þ dkn ek
uþ1
n pi dj1 are all distinct for j ¼ u þ 1; . . . ; k: Note also that
k Y
pi djn 5
jþ1
Qt
1
Qk u
pi
qpj log pj log2 pj ðlog3 pj Þ5 Qt
5
si
1
si
Qk u
pi
2
pk ðlog pk Þ ðlog2 pk Þ2
P for j ¼ u; u þ 1; . . . ; Q k 1:Q Thus to show r¼ di ; where di are Q Q t k distinct divisors of and di 5 t1 si ku pi =pk ðlog pk Þ2 ðlog2 pk Þ2 ; 1 si u pi P it suffices to showQ rku ¼ di0 ; where di0 are distinct divisors Qt t 2 2 0 of 1 si and di 5 1 si =pk ðlog pk Þ ðlog2 pk Þ : Recall that our rku satisfies rku
! t ðlog2 pk Þ2 Y ¼ ð1 þ oð1ÞÞ si : log pk 1
P 0 Then 2, we have rku ¼ di ; whereQdi0 are distinct divisors of Q Qt by Lemma t t 2 0 1 si and di 5 1 si =pkðtÞ log pkðtÞ ð1 log pkðtÞ Þ5 1 si =LðtÞ: ] 2
Proof of Lemma 4. We note that every integer d4st is in DðtÞ and by the choice of pkðtÞ ; every prime popkðtÞ is in DðtÞ: Thus for deDðtÞ and d4LðtÞ; d must reside in one of the following sets: D1 ¼ fd4LðtÞ: there exists m2 such that st om2 oLðtÞ and m2 j dg; D2 ¼ fd4LðtÞ: there exists p such that pkðtÞ opoLðtÞ and p j dg:
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
365
Then X X 1 1 X 1 X 1 4 4 þ : d d4LðtÞ;deDðtÞ d d2D d d2D d d2D 2
We first estimate X 1 4 d d2D 1
1
2
P
1 d2D1 d :
X st om2 4LðtÞ
1 m2
2 ½LðtÞ=m X
j¼1
1 j
4 ðlog LðtÞ log sðtÞ þ 1Þ
X st om2 oLðtÞ
4 ðlog LðtÞ log sðtÞ þ 1Þ
X
1 m2
pffiffiffiffiffiffi pffiffiffi st omo LðtÞ
1 1 m1 m
! log pkðtÞ log pkðtÞ ¼O ¼O : pffiffiffiffi st ðpkðtÞ Þ1=4 P We next estimate d2D2 d1: To do this, we break the interval ½pkðtÞ ; LðtÞ into three subintervals of the form ! " ! " LðtÞ LðtÞ LðtÞ LðtÞ ; LðtÞ : pkðtÞ ; ; ; ; log LðtÞ ðlog LðtÞÞ2 ðlog LðtÞÞ2 log LðtÞ Let 1
QðmÞ ¼ LðtÞðlog LðtÞÞm ; RðmÞ ¼ LðtÞðlog LðtÞÞ
mþ1 m ;
PðmÞ ¼ fp: QðmÞopoQðm þ 1Þg; RðmÞ ¼ fp: RðmÞopoRðm þ 1Þg: Then X 1 ¼ d d2D 2
¼
½LðtÞ=p 1 X 1 p j¼1 j opoLðtÞ
X
pkðtÞ
X pkðtÞ opoR1
¼ S1 þ S2 ;
0 1 ½LðtÞ=p 1 X 1 X 1 ½LðtÞ=p X 1 X 1 ½LðtÞ=p 1 X 1 X @ A þ þ p j¼1 j m¼1 p2RðmÞ p j¼1 j p2PðmÞ p j¼1 j say:
366
HISASHI YOKOTA
We shall estimate S1 and S2 : Note that by [7], Oð1=log2 xÞ: Then we have S1 ¼
X pkðtÞ opoR1
P
1 p4x p
¼ log2 x þ B þ
½LðtÞ=p X 1 X 1 1 4ðlog LðtÞ log pkðtÞ þ 1Þ p j¼1 j p pkðtÞ opoR1 t
1 ¼ ðlog LðtÞ log pkðtÞ þ 1Þ log2 Rð1Þ log2 pkðtÞ þ O 2 log pkðtÞ
!!
Since LðtÞ ¼ pkðtÞ ðlog pkðtÞ Þ2 ðlog2 pkðtÞ Þ2 ; we have LðtÞ Rð1Þ ¼ 4pkðtÞ ðlog2 pkðtÞ Þ2 : 2 ðlog LðtÞÞ Thus,
! log3 pkðtÞ ðlog2 pkðtÞ Þ2 S1 ¼ ð2 þ oð1ÞÞ log2 pkðtÞ O ¼o : log pkðtÞ log pkðtÞ
For the estimate of S2 ; we note that 1 X 1 ðm þ 1Þ log2 LðtÞ k log2 RðmÞ ¼ log2 LðtÞ ; k m log LðtÞ k¼1 log2 QðmÞ ¼ log2 LðtÞ
1 X 1 log2 LðtÞ k : k m log LðtÞ k¼1
Then, X 1 1 ¼ log2 Rðm þ 1Þ log2 RðmÞ þ O 2 p log RðmÞ p2RðmÞ
!
! 1 X 1 mþ1 k mþ2 k log2 LðtÞ k ¼ k m mþ1 log LðtÞ k¼1 1 þ O 2 log RðmÞ
!
! 1 log2 LðtÞ 1 log2 LðtÞ 2 þO ¼ mðm þ 1Þ log LðtÞ m2 ðm þ 1Þ log LðtÞ
1 þO 2 log RðmÞ
!
:
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
and X 1 1 ¼ log2 Qðm þ 1Þ log2 QðmÞ þ O 2 p log QðmÞ p2PðmÞ
367
!
! k k ! 1 X 1 1 1 log2 LðtÞ k 1 ¼ þO k m mþ1 log LðtÞ log2 QðmÞ k¼1 ! 1 1 log2 LðtÞ 1 log2 LðtÞ 2 þO ¼ m m þ 1 log LðtÞ m2 ðm þ 1Þ log LðtÞ
! 1 þ O : log2 QðmÞ Thus, X 1 X 1 X 1 ½LðtÞ=p 4 p j¼1 j p2RðmÞ p p2RðmÞ
mþ1=m ½ðlog LðtÞÞ
X
j¼1
mþ1 log2 LðtÞ þ 1 4 m
¼
1 j
X 1 p p2RðmÞ
1 ðlog2 LðtÞÞ2 log2 LðtÞ þ O m2 log LðtÞ mðm þ 1Þ log LðtÞ log2 LðtÞ þO log2 RðmÞ
!
and 1=m
LðtÞÞ X 1 ½LðtÞ=p X 1 X 1 ½ðlogX 1 4 p j¼1 j p2PðmÞ p j j¼1 p2PðmÞ X 1 1 log LðtÞ þ 1 4 m 2 p p2PðmÞ 1 1 ðlog2 LðtÞÞ2 ¼ m2 mðm þ 1Þ log LðtÞ
þ O
! log2 LðtÞ log2 LðtÞ þO : mðm þ 1Þ log LðtÞ log2 QðmÞ
368
HISASHI YOKOTA
Now we evaluate S2 : 0 1 1 X 1 X 1 ½LðtÞ=p X 1 X 1 ½LðtÞ=p X @ A þ S2 ¼ p j p j j¼1 j¼1 m¼1 p2RðmÞ p2PðmÞ 1 X 2 1 ðlog2 LðtÞÞ2 4 m2 mðm þ 1Þ log LðtÞ m¼1 !! log2 LðtÞ log2 LðtÞ þ O þO : mðm þ 1Þlog LðtÞ log2 RðmÞ
We note that 1 X 2 1 2p2 1; ¼ 2 m mðm þ 1Þ 6 m¼1 ! 1 1 1 dx ¼ O ¼O : 2 log Rð1Þ log LðtÞ log Rð1Þ x
Z
1 X
1 ¼O 2 m¼1 log RðmÞ
1
Thus, 2 2 ðlog2 pkðtÞ Þ2 2p 6 ðlog2 LðtÞÞ2 p 3 þ oð1Þ þ oð1Þ S2 4 ¼ : 6 3 log LðtÞ logpkðtÞ This proves that for t large, X 1 p2 3 ðlog2 pkðtÞ Þ2 4 þ oð1Þ : d 3 log pkðtÞ d4LðtÞ deDðtÞ
On the other hand, X 1 X 1 5 5S2 d d2D d d4LðtÞ 2
deDðtÞ
and S2 ¼
1 X m¼1
0
1 X 1 X 1 ½LðtÞ=p X 1 X 1 ½LðtÞ=p @ A: þ p j¼1 j p2PðmÞ p j¼1 j p2RðmÞ
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
369
Since X 1 X 1 X 1 ½LðtÞ=p 5 p j¼1 j p2RðmÞ p p2RðmÞ
mþ2=ðmþ1Þ ½ðlog LðtÞÞ
X
j¼1
1 j
X mþ2 1 1 log2 LðtÞ þ g þ O ¼ mþ1 log LðtÞ p p2RðmÞ ðlog2 LðtÞÞ2 log2 LðtÞ ¼ þO mðm þ 1Þ log LðtÞ mðm þ 1Þ2 log LðtÞ mþ2
1 þ O 2 log RðmÞ
!
and X 1 ½LðtÞ=p X 1 X 1 5 p j¼1 j p2PðmÞ p p2PðmÞ
1=ðmþ1Þ ½ðlog LðtÞÞ
X
j¼1
1 j
¼
X 1 1 1 log2 LðtÞ þ g þ O mþ1 log LðtÞ p p2PðmÞ
¼
ðlog2 LðtÞÞ2 log2 LðtÞ þ O mðm þ 1Þ log LðtÞ mðm þ 1Þ2 log LðtÞ 1
! 1 þO ; log2 QðmÞ we have S2 5
1 X m¼1
¼
1 X m¼1
! ! ðlog2 LðtÞÞ2 log2 LðtÞ þO þ mðm þ 1Þ log LðtÞ log LðtÞ mðm þ 1Þ2 mðm þ 1Þ2 mþ2
1
m þ 3 ðlog2 LðtÞÞ2 mðm þ 1Þ2 log LðtÞ
! þO
log2 LðtÞ : log LðtÞ
We note that mþ3 mðm þ 1Þ
2
¼
3ðm þ 1Þ 2m mðm þ 1Þ
2
¼
3 2 : mðm þ 1Þ ðm þ 1Þ2
370
HISASHI YOKOTA
Thus,
S2 5
1 X m¼1
! 3 2 ðlog2 LðtÞÞ2 log2 LðtÞ þO mðm þ 1Þ ðm þ 1Þ2 log LðtÞ log LðtÞ
2 p ðlog2 LðtÞÞ2 ¼ 32 1 þ oð1Þ 6 log LðtÞ ¼
ðlog2 pkðtÞ Þ2 15 p2 þ oð1Þ : 3 log pkðtÞ
]
Proof of Lemma 5. We first note that for st prime fs1 ; s2 ; . . . ; st ; puðtÞ ; . . . ; pkðtÞ g ¼ fs1 ; s2 ; . . . ; st1 ; puðt1Þ ; . . . ; pkðt1Þ g [ fpkðt1Þþ1 ; . . . ; pkðtÞ g i
and for st ¼ p2 for some i51; fs1 ; s2 ; . . . ; st ; puðtÞ ; . . . ; pkðtÞ g ¼ fs1 ; s2 ; . . . ; st1 ; puðt1Þ ; . . . ; pkðt1Þ g [ fst ; pkðt1Þþ1 ; . . . ; pkðtÞ g:
Thus in either case, we have X 1 X 1 X 1 X ¼ d d4Lðt1Þ d d4LðtÞ d d4LðtÞ d4LðtÞ d2DðtÞ
d2Dðt1Þ
d2DðtÞ
þ
d2Dðt1Þ
X d4LðtÞ d2Dðt1Þ
4
1 d
X 1 1 d d4Lðt1Þ d d2Dðt1Þ
½LðtÞ=s ½LðtÞ=p LðtÞ X X 1 Xt 1 1 X 1 1 þ þ st j¼1 j pkðt1Þ op4pkðtÞ p j¼1 j j¼Lðt1Þ j
¼ S1 þ S2 þ S3 ;
say:
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
371
Now we estimate S1 ; S2 ; S3 : Since by Lemma 1, log pkðt1Þ 5log pkðtÞ 1 O 1=5 ; we have pkðtÞ ! ½LðtÞ=s log pkðtÞ 1 Xt 1 1 4 ðlog LðtÞ log st þ 1Þ ¼ O pffiffiffiffiffiffiffiffiffi ; S1 ¼ st j¼1 j st pkðtÞ X
S2 ¼
pkðt1Þ op4pkðtÞ
½LðtÞ=p X 1 X 1 1 4ðlog LðtÞ log pkðt1Þ þ 1Þ p j¼1 j p pkðt1Þ op4pkðtÞ
¼ ð2 þ oð1ÞÞ log2 pkðtÞ
2 log2 pkðtÞ log2 pkðt1Þ þ 2 log pkðt1Þ
!
log2 pkðtÞ ¼ ð4 þ oð1ÞÞ 2 ; log pkðtÞ S3 ¼
LðtÞ X j¼Lðt1Þ
1 1 ¼ log LðtÞ log Lðt 1Þ þ O j Lðt 1Þ
0
1 1 ¼ O@ 1=5 A: pkðtÞ Thus, we have X 1 X 1 log2 pkðtÞ 4ð4 þ oð1ÞÞ 2 : d d log pkðtÞ d4LðtÞ d4Lðt1Þ d2DðtÞ
d2Dðt1Þ
This completes the proof of Lemma 5.
]
REFERENCES 1. M. N. Bleicher, P. Erdo+ s, Denominators of Egyptian fractions, II, Illinois J. Math. 20 (1976), 598–613. 2. E. S. Croot III, On some question of Erdo+ s and Graham about Egyptian fractions, Mathematika 46 (1999), 359–372. 3. H. Davenport, On the addition of residue classes, J. London Math. Soc. 10 (1935), 30–32. 4. D.R. Heath-Brown and H. Iwaniec, On the difference between consecutive primes, Inv. Math. 55 (1979), 49–69. 5. P. Erdo+ s and R. L. Graham, ‘‘Old and New Problems and Results in Combinatorial Number Theory,’’ pp. 30–44. Monographie No. 28, L’Enseign. Math. Univ. de Geneve`, 1980.
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6. G. Lorentz, On a problem of additive number theory, Proc. Amer. Math. Soc. 5 (1954), 838–841. 7. J. Rosser and L. Schoenfeld, Approximate formula for some functions of prime numbers, Illinois J. Math. 6 (1962), 64–94. 8. H. Yokota, Denominators of Egyptian fractions, J. Number Theory 28 (1988), 258–271. 9. H. Yokota, On a problem of Erdo+ s and Graham, J. Number Theory 39 (1991), 327–338. 10. H. Yokota, On number of integers representable as sums of unit fractions, II, J. Number Theory 67 (1997), 162–169. 11. H. Yokota, Corrigendum, J. Number Theory 72 (1998), 150. 12. H. Yokota, The largest integer expressible as a sum of reciprocal of integers, J. Number Theory 76 (1999), 206–216. 13. H. Yokota, Erratum, J. Number Theory 83 (2000), 183–184.
On the Number of Integers Representable as Sums of Unit Fractions, III Hisashi Yokota Department of Mathematics, Hiroshima Institute of Technology, 2-1-1 Miyake Saeki-ku, Hiroshima, Japan E-mail: [email protected] Communicated by D. Goss Received July 16, 2001; revised January 4, 2002
Let NðnÞ be the set of all integers that can be expressed as a sum of reciprocals of distinct integers 4n: Then we prove that for sufficiently large n; log n þ g
2 p ðlog2 nÞ2 þ oð1Þ 4jNðnÞj; 3 log n
which improves the lower bound given by Croot.
1.
# 2002 Elsevier Science (USA)
INTRODUCTION
Let NðnÞ be the set of all integers a that can be expressed as a¼
X ek ; k 14k4n
where ek takes either 0 or 1: Answering questions of Erdo+ s and Graham [5], we have shown (see [10–13]) that 1 ðlog2 nÞ2 þ oð1Þ : log n þ g 2 oð1Þ4jNðnÞj4log n þ g 4 log n Then recently, in [2], Croot improved our previous results so that
9 ðlog2 nÞ2 1 ðlog2 nÞ2 þ oð1Þ þ oð1Þ log n þ g 4jNðnÞj4log n þ g : 2 2 log n log n Here and in the sequel, we let logj denote the jth fold iterated logarithm. 351 0022-314X/02 $35.00 # 2002 Elsevier Science (USA) All rights reserved.
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HISASHI YOKOTA
In connection with NðnÞ; Don Zagier, in private communication, pointed out the following: Let a be a given integer and define F ðaÞ ¼ minfn : a 2 NðnÞg: Then determining NðnÞ for all n is the same as calculating F ðaÞ since a 2 NðnÞ iff n5F ðaÞ: He then asked if it is possible to improve the upper bound of F ðaÞ: Let S :¼ fsj : j51g be the increasing sequence of all positive integers of i the form p2 ; i50; where p a prime. Once st is chosen, we denote pkðtÞ the largest prime os2t ; puðtÞ the smallest prime > st : We define 8 9 t kðtÞ < = Y Y DðtÞ ¼ d4LðtÞ: d si pj ; : ; 1 uðtÞ where LðtÞ ¼ pkðtÞ ðlog pkðtÞ Þ2 ðlog2 pkðtÞ Þ2 : Now we ask how many integers a can be expressed as a¼
X
ek ; k k2DðtÞ;k4n
where ek takes either 0 or 1: Let N n ðnÞ be the set of all integers a that can be expressed as a¼
X k2DðtÞ;k4n
ek ; k
where ek takes either 0 or 1 and F n ðaÞ ¼ minfn: a 2 N n ðnÞg: Then N n ðnÞ NðnÞ and jN n ðnÞj4jNðnÞj: In this paper, we prove the following results. Theorem 1.
There exists a constant n0 such that, for all n > n0 ;
p2 ðlog2 nÞ2 þ oð1Þ 4jN n ðnÞj log n þ g 3 log n and "
# p2 ðlog aÞ2 F ðaÞ4exp a g þ þ oð1Þ : 3 a n
By noting jN n ðnÞj4jNðnÞj and F ðaÞ4F n ðaÞ; we improve the lower bound of jNðnÞj and the upper bound of F ðaÞ: Also by quoting Croot’s result about
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
353
the upper bound of jNðnÞj; we obtain Corollary 1. There exists a constant n0 such that, for all n > n0 ; p2 ðlog2 nÞ2 1 ðlog2 nÞ2 þ oð1Þ þ oð1Þ 4jNðnÞj4log n þ g log n þ g 2 3 log n log n
and "
# 2 p ðlog aÞ2 þ oð1Þ F ðaÞ4F ðaÞ4exp a g þ : 3 a n
2. Lemma 1.
LEMMATA 2=5
For all large pkðtÞ ; pkðt1Þ 5pkðtÞ ð1 21=5 Þ: pkðtÞ
Lemma 2. There exists a constant t0 such that, for all t5t0 and ðlog2 pk Þ2 Qt all r satisfying r ¼ ð1 þ oð1ÞÞ there are distinct 1 si ; Q P log pk Q integers di with the property that r ¼ di ; where di j t1 si and di 5 t1 si =pk log pk ð1 log2 pk Þ: 2
Lemma 3. There exists a constant2 Q t0 such Qk that, for all t5t0 and t 2 pk Þ all r satisfying r ¼ ð1 þ oð1ÞÞðlog s 1 i u pj ; QthereQ are distinct log pP k integers d with the property that r ¼ d ; where di j t1 si ku pj and di 5 i Qt Qk i 1 si u pj =LðtÞ: Lemma 4.
There exists a constant t0 such that for all t5t0
2 X ðlog2 pkðtÞ Þ2 ðlog2 pkðtÞ Þ2 15 p2 1 p 3 4 þ oð1Þ þ oð1Þ 4 : d 3 3 log pkðtÞ log pkðtÞ d4LðtÞ;deDðtÞ
Lemma 5.
There exists a constant t0 such that for all t5t0 X 1 X 1 log2 pkðtÞ 4ð4 þ oð1ÞÞ 2 : d d log pkðtÞ d4LðtÞ d4Lðt1Þ d2DðtÞ
d2Dðt1Þ
Assuming these lemmas hold, we prove Theorem 1.
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HISASHI YOKOTA
3.
PROOF OF THEOREM 1
Given a large integer a; we choose t such that X d4Lðt1Þ d2Dðt1Þ
X 1 1 ½ðlog aÞ2 oa þ o : d a d d4LðtÞ d2DðtÞ
Since by Lemmas 4 and 5,
aþ
LðtÞ X 1 X X 1 ½ðlog aÞ2 1 o ¼ a d j d4LðtÞ d j¼1 d4LðtÞ d2DðtÞ
deDðtÞ
ðlog2 pkðtÞ Þ2 15 p2 þ oð1Þ 4 log LðtÞ þ g 3 log pkðtÞ and
aþ
X 1 X 1 log2 pkðtÞ ½ðlog aÞ2 > 5 ð4 þ oð1ÞÞ 2 a d d4LðtÞ d log pkðtÞ d4Lðt1Þ d2Dðt1Þ
d2DðtÞ LðtÞ X 1 X log2 pkðtÞ 1 ð4 þ oð1ÞÞ 2 ¼ j d log pkðtÞ j¼1 d4LðtÞ deDðtÞ
5
2 LðtÞ X ðlog2 pkðtÞ Þ2 1 p 3 þ oð1Þ j 3 log pkðtÞ j¼1
2 ðlog2 pkðtÞ Þ2 p 3 þ oð1Þ ¼ log LðtÞ þ g : 3 log pkðtÞ 2
Let a0 ¼ log LðtÞ þ g ðp3 þ oð1ÞÞ ðlog2 pkðtÞ Þ2 ; we have
ðlog2 pkðtÞ Þ2 log pkðtÞ :
Then since LðtÞ ¼ pkðtÞ ðlog pkðtÞ Þ2
2 ðlog2 pkðtÞ Þ2 ½ðlog a0 Þ2 p 3 þ oð1Þ a0 þ ¼ log LðtÞ þ g : a0 3 log pkðtÞ
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
355
2 ðlog2 pkðtÞ Þ2 p þ oð1Þ a5a0 ¼ log LðtÞ þ g : 3 log pkðtÞ
ð1Þ
Thus,
On the we have
other
hand,
for
2
b0 ¼ log LðtÞ þ g ð18p 3 þ oð1ÞÞ
ðlog2 pkðtÞ Þ2 log pkðtÞ ;
ðlog2 pkðtÞ Þ2 ½ðlog b0 Þ2 15 p2 b0 þ þ oð1Þ ¼ log LðtÞ þ g : b0 3 log pkðtÞ Thus, a4b0 ¼ log LðtÞ þ g
ðlog2 pkðtÞ Þ2 18 p2 þ oð1Þ : 3 log pkðtÞ
This shows that a ¼ ð1 þ oð1ÞÞ log pkðtÞ : Now we show that X ej a¼ : j j4LðtÞ j2DðtÞ
Since a ¼ ð1 þ oð1ÞÞ log pkðtÞ ; a 2 DðtÞ; and by Lemma 5, we have X 1 d d4LðtÞ
! log2 pkðtÞ ½ðlog aÞ2 r aþ : ¼ Qt QkðtÞ 4ð4 þ oð1ÞÞ 2 a log pkðtÞ 1 si uðtÞ pj
d2DðtÞ
Rewriting this, we have 0 1 2 X 1 r ½ðlog aÞ
A @Qt QkðtÞ þ a¼ d a s p d4LðtÞ 1 i uðtÞ j d2DðtÞ
0 1 n X 1 r ¼ @Qt QkðtÞ A; d d4LðtÞ 1 si uðtÞ pj d2DðtÞ
where rn satisfies the following: rn ¼ ð1 þ oð1ÞÞ
kðtÞ t Y ðlog2 pkðtÞ Þ2 Y si pj : log pkðtÞ 1 uðtÞ
ð2Þ
356
HISASHI YOKOTA
Then by Lemma 3, we can express rn as a sum of distinct divisors Qt QkðtÞ si pj Q P QkðtÞ di of t1 si uðtÞ pj with di 5 1 LðtÞ uðtÞ such that rn ¼ m 1 di : Thus, Pm X 1 X ej 1 di Qt Q a¼ : ¼ kðtÞ d j d4LðtÞ j4LðtÞ 1 si uðtÞ pj d2DðtÞ
j2DðtÞ
This shows that for every large integer a; we can find t so that a is a sum of reciprocals of distinct integers in DðtÞ less than or equal to LðtÞ; where LðtÞ satisfies 2 ðlog2 pkðtÞ Þ2 p þ oð1Þ a5log LðtÞ þ g 3 log pkðtÞ and ðlog2 pkðtÞ Þ2 18 p2 þ oð1Þ : a4log LðtÞ þ g 3 log pkðtÞ We note that a¼
X ej j j4LðtÞ j2DðtÞ
implies that a 2 N n ðLðtÞÞ ð1 þ oð1ÞÞ log pkðtÞ ; we have
and
F n ðaÞ4LðtÞ:
Then
since
a¼
2 ðlog2 pkðtÞ Þ2 p2 p ðlog aÞ2 þ oð1Þ þ oð1Þ : a5log LðtÞ þ g ¼ log LðtÞ þ g 3 log pkðtÞ 3 a
Thus, "
# p2 ðlog aÞ2 F ðaÞ4LðtÞ4exp a g þ þ oð1Þ : 3 a n
We also note that a 2 N n ðLðtÞÞ implies
ðlog2 pkðtÞ Þ2 p2 þ oð1Þ log LðtÞ þ g 4jN n ðLðtÞÞj: 3 log pkðtÞ
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
357
Let n be an integer such that LðtÞ4noLðt þ 1Þ: Then log LðtÞ4log n olog Lðt þ 1Þ and by Lemma 1 log Lðt þ 1Þ ¼ ð1 þ oð1ÞÞ log pkðtþ1Þ 0
0
11 1 ¼ ð1 þ oð1ÞÞ@log pkðtÞ þ O@ 1=5 AA pkðtÞ 0 1 1 ¼ log LðtÞ þ O@ 1=5 A: pkðtÞ
Therefore, log n þ g
4.
2 p ðlog2 nÞ2 þ oð1Þ 4jN n ðnÞj: 3 log n
PROOF OF LEMMAS 3=5
Since for pi large, by [4], we have piþ1 4pi þ pi :
Proof of Lemma 1.
3=5
6=5
Thus pkðt1Þ 5pkðt1Þþ1 pkðt1Þ 5s2t1 st1 : By the choice of st ; we have 3=5
st1 5st st1 : This then implies that pkðt1Þ 5ðst
¼ s2t 1
3=5 st1 Þ2
22=5
!
4=5
st
6=5 st 5s2t
0 5pkðtÞ @1
8=5 2st 5s2t
1 22=5 A 1=5
:
2=5 ! 2 1 st
]
pkðtÞ
l
Proof of Lemma 2. Case 1: st ¼ p2 ; l51: Define for j ¼ 1; 2; 3; . . . ; 2l ; (Q ) t1 Y t1 si j j 1 : p p þ 14d4p 1; d Dj ¼ s [ f0g: 1 i p2l j d We note that if dj 2 Dj and dj a0; then for all i; Qt1
Qt1 si si 4dj 4 2l j j1 : l j 2 j p ðp 1Þ p ðp p þ 1Þ 1
358
HISASHI YOKOTA
Q Q Q 2l 1 Then by the property of S; t1 ti ; where p[ ti for all i: 1 si ¼ p Thus, (Q ) t1 Y t1 si j j 1 : p p þ 14d4p 1; d s [ f0g Dj ¼ 1 i p2l j d ( ¼ pj1
t1 ) Y : pj p þ 14d4pj 1; d si [ f0g d 1
Q
ti
f0; pj1 ; 2pj1 ; . . . ; ðp 1Þpj1 g ðmod pj Þ: Pl Let Dn ¼ 21 Dj : Then by Lorentz’s theorem [6], Dn is a complete n residue system P l modulo st : Thus given any integer r; r dt ðmod st Þ for some dtn ¼ 21 dj ej 2 Dn ; where dj 2 Dj ; and dtn 4
t1 Y
si
2l X
1
1
1 p2l j ðpj p þ 1Þ
!
t1 Y 2 log st si 4 : st log 2 1 Case 2: st ¼ p: Define (Q D¼
) Y t1 si st 1 4dost ; d : s [ f0g: 1 i 2 d
t1 1
Then by Lemma 1 of [9], jDj ¼ pþ3 2 and no two elements are congruent modulo st : If r 2d ðmod st Þ for some d 2 D; then let Dn ¼ ðD fdgÞ þ ðD fdgÞ: Otherwise, Dn ¼ D þ D:PThen by Cauchy–Davenport Theorem [3], r dtn ðmod st Þ for some dtn ¼ 21 dj ej 2 Dn ; where dj 2 D; and n
dt 4
t1 Y 1
4
t1 Y 1
2 2 þ si st 1 st þ 1
2 log st si : st log 2
Thus in either case, given any integer r; we have r dtn ðmod st Þ and dtn 4
t1 Y 1
si
2 log st : st
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
359
Now given r such that r ¼ ð1 þ oð1ÞÞ
t ðlog2 pkðtÞ Þ2 Y si : log pkðtÞ 1
We let r1 ¼ Then
r dtn ; st
an integer:
! t1 ðlog2 pkðtÞ Þ2 2 log st Y r1 5 ð1 þ oð1ÞÞ 2 si log pkðtÞ st log 2 1
and r1 4ð1 þ oð1ÞÞ
t1 ðlog2 pkðtÞ Þ2 Y si : log pkðtÞ 1
Continue this process j times so that stj is the smallest element in S satisfying log pkðtÞ ð1 log 2pkðtÞ Þ4stj : Then, 2 ! tj t ðlog2 pkðtÞ Þ2 2 X log si Y rj 5 ð1 þ oð1ÞÞ si : 2 log 2 tj si log pkðtÞ 1 Since Z 2pffiffiffiffiffiffi t pkðtÞ X log si 4 s2i log pkðtÞ ð1 tj
2 log2 pkðtÞ
2pffiffiffiffiffiffi pkðtÞ
log x 1 ¼ x x ¼ ð1 þ oð1ÞÞ
log2 pkðtÞ log x dx þ O 2 ðlog pkðtÞ Þ2 Þ x þO
log pkðtÞ ð12=log2 pkðtÞ Þ
!
log2 pkðtÞ
!
ðlog pkðtÞ Þ2
log2 pkðtÞ ; log pkðtÞ
! tj ðlog2 pkðtÞ Þ2 log2 pkðtÞ Y 2 ð1 þ oð1ÞÞ rj 5 ð1 þ oð1ÞÞ si log 2 log pkðtÞ log pkðtÞ 1 ! tj ðlog2 pkðtÞ Þ2 Y ¼ ð1 þ oð1ÞÞ si : log pkðtÞ 1
360
HISASHI YOKOTA
We also note that tj ðlog2 pkðtÞ Þ2 Y rj 4ð1 þ oð1ÞÞ si : log pkðtÞ 1
Thus, rj ¼ ð1 þ oð1ÞÞ
tj ðlog2 pkðtÞ Þ2 Y si : log pkðtÞ 1
Q Since by Lemma 1 of [8], every integer a4 tj si can be expressed as a sum 1 Qtj of distinct divisors of 1 si : Thus, we can express rj as follows: tj Y X rj ¼ di0 ; where di0 si ; di0 51: 1
Then r ¼ st r1 þ dtn n ¼ st ðst1 r2 þ dt1 Þ þ dtn
.. . ¼
!
t Y
si rj þ
tjþ1
¼
t Y tjþ1
si
!
t Y
n si dtjþ1 þ
tjþ2
X
di0 þ
!
t Y
n si dtjþ2 þ þ dtn
tjþ3 t Y
! n si dtjþ1 þ
tjþ2
t Y
! n si dtjþ2 þ þ dtn
tjþ3
Q Q n and ttmþ1 si dtm are allPdistinct for m ¼ 0; 1; 2; . . . ; j: Here ttþ1 si ¼ 1: s n Note that for dtm ¼ 1tm dj ej a0; we have by Cases 1 and 2, Qtm1 1 dj 5stm 1 si : Thus, t t tm1 Y Y 1 Y s i dj > si si stm1 tmþ1 tmþ1 1 Qt 5
Qt si 1 si 5 1 : 2 st 4pkðtÞ
Q Since stj1 4log pkðtÞ ð1 log 2pkðtÞ Þ; we have tj1 si 4pkðtÞ by Lemma 2 of 1 2 [1]. Thus, Qt Qt Qt t Y 1 si 1 si 1 si : si ¼ Qtj ¼ Qtj1 5 si si stj pkðtÞ log pkðtÞ ð1 log22pkðtÞ Þ tjþ1 1 1
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
361
P Q This Qshows that r ¼ di ; where di are distinct divisors of t1 si and di 5 t1 si =pkðtÞ log pkðtÞ ð1 log 2pkðtÞ Þ: ] 2
For u ¼ uðtÞ4j4k ¼ kðtÞ; let
Proof of Lemma 3. ( Dj ¼
) Y t d: 14d4log pj log2 pj ðlog3 pj Þ ; d s : 1 i 5
Let q be the smallest element in S satisfying q > log pkðtÞ : Then qo2 log pkðtÞ : Now define for j ¼ u; u þ 1; . . . ; k (Q
n
Dj ¼ where
Qu1 u
t 1
Qj1
si
u
pi
qd
) : d 2 Dj ;
pi ¼ 1: Note that if djn 2 Dnj ; then djn is an integer and Qt
1
si
Qj1 u
pi
n
q log pj log2 pj ðlog3 pj Þ
5
Qt
4dj 4
1
si
Qj1 u
pi
q
:
We claim that (
mj X
) djn ej : ej ¼ 0 or 1
f1; 2; . . . ; pj g ðmod pj Þ;
1
where mj 4q log pj ðlog2 pj Þ2 : Let a be a residue modulo pj : Choose tn such that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi log pj ð1 2=log2 pj Þ4stn 42 log pj ð1 2=log2 pj Þ: Then by Lemma 1 of [8], tn Y 1
Q tn 1
si
si opj : Thus, we can choose kn so that
n kY 1
un
pi opj o
tn Y 1
si
kn Y
pi ;
un
where pun is the smallest prime greater than stn : Then by Lemma 2 of [1], pffiffiffiffiffiffiffi log pj ð1 2=log2 pj Þ4pkn 4log pj ð1 þ 2=log2 pj Þ: Thus stn =24 pkn 42stn : Now, consider Qn Q n pj s þ rn a a t si kn pi ¼ Q1tn Qukn ¼ Q tn Q k n ; pj pj 1 si un pi pj 1 si un pi
362
HISASHI YOKOTA
where rn is chosen so that t k Y pffiffiffiffiffi Y ð1 þ 1=log2 pkn 2= stn Þ si pi n
n
1
un
t k Y pffiffiffiffiffi Y 4rn oð2 þ 1=log2 pkn 2= stn Þ si pi : n
n
1
un
P
n Then by [9], di ; where di are distinct divisors of QknTheorem 2.2Qof Qkrn ¼ Q tn tn 5 n log pk n ðlog pk n Þ : Thus, s p and d 5 s p =p n n i i i i i k 2 1 u 1 u
a
tn Y
si
1
kn Y
pi r n
un
tn Y
si
kn Y
1
un
tn Y
kn Y
si
1
!
P p i Q tn 1
pi
un
si
mj X ei i i¼1
di Qk n un
pi
! ðmod pj Þ;
Q tn Q k n where mj is less than or equal to pkn log pkn ðlog2 pkn Þ5 : Since 1 nsi un pi ; Q tn ð Q pj Þ ¼ 1; as a runs through all residues modulo pj ; a 1 si kun pi runs through all residues modulo pj : Thus, n ) ( n ! mj Y t kn k Y Y X si pi ei =i : ei ¼ 0 or 1; i p un i 1 i¼1 un contains all residues modulo pj : Furthermore, t Y 1
si
j1 Y u
pi =q
tn Y 1
si
kn Y
! pi ; pj
¼ 1:
un
Thus, we have mj X
djn ej f1; 2; . . . ; pj g ðmod pj Þ;
1
where mj 4qpkn log pkn ðlog2 pkn Þ5 4q log pj ðlog2 pj Þ2 for pj large. Now given r such that ! t k Y ðlog2 pk Þ2 Y r ¼ ð1 þ oð1ÞÞ si pj ; log pk u 1
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
363
we can express r as r
X
dkn ek ðmod pk Þ;
where dkn 2 Dnk
and X
Let r1 ¼ ðr
! mk t k 1 t k 1 Y X Y 1Y ei log mk Y dk e k ¼ si pi si pi : 4 q 1 i q u u i¼1 1 n
P
dkn ek Þ=pk : Then,
r1 5 ð1 þ oð1ÞÞ
t k 1 t k 1 Y Y ðlog2 pk Þ2 Y log mk Y si pi si pi log pk qpk 1 u u 1
ðlog2 pk Þ2 log mk ¼ ð1 þ oð1ÞÞ log pk qpk
!
t Y
si
k 1 Y
1
pi
u
and r1 4
t k 1 Y r ðlog2 pk Þ2 Y ¼ ð1 þ oð1ÞÞ si pi : pk log pk u 1
Repeating the same argument k u times, we have k ðlog2 pk Þ2 log mk X 1 rku 5ð1 þ oð1ÞÞ p log pk q j¼u j
!
t Y
si :
1
2 Since mk oq Pklog 1pk ðlog pk Þ ; we have log mk 4ð2 þ oð1ÞÞ log2 pk : We note that since j¼u p ¼ Oð1Þ;
k log mk X 1 log2 pk ¼O : p q log pk j¼u j Thus, we have ! Y t ðlog2 pk Þ2 log2 pk O si rku 5 ð1 þ oð1ÞÞ log pk log pk 1 ¼ ð1 þ oð1ÞÞ
t ðlog2 pk Þ2 Y si : log pk 1
364
HISASHI YOKOTA
Therefore, rku ¼ ð1 þ oð1ÞÞ
t ðlog2 pk Þ2 Y si : log pk 1
Note that X r ¼ pk r 1 þ dkn ek X X n ¼ pk pk1 r2 þ dk1 ek þ dkn ek ¼
k Y u
and
Qk j
pi rku þ
k Y
pi
X
X X n dun eu þ þ pk dk1 ek1 þ dkn ek
uþ1
n pi dj1 are all distinct for j ¼ u þ 1; . . . ; k: Note also that
k Y
pi djn 5
jþ1
Qt
1
Qk u
pi
qpj log pj log2 pj ðlog3 pj Þ5 Qt
5
si
1
si
Qk u
pi
2
pk ðlog pk Þ ðlog2 pk Þ2
P for j ¼ u; u þ 1; . . . ; Q k 1:Q Thus to show r¼ di ; where di are Q Q t k distinct divisors of and di 5 t1 si ku pi =pk ðlog pk Þ2 ðlog2 pk Þ2 ; 1 si u pi P it suffices to showQ rku ¼ di0 ; where di0 are distinct divisors Qt t 2 2 0 of 1 si and di 5 1 si =pk ðlog pk Þ ðlog2 pk Þ : Recall that our rku satisfies rku
! t ðlog2 pk Þ2 Y ¼ ð1 þ oð1ÞÞ si : log pk 1
P 0 Then 2, we have rku ¼ di ; whereQdi0 are distinct divisors of Q Qt by Lemma t t 2 0 1 si and di 5 1 si =pkðtÞ log pkðtÞ ð1 log pkðtÞ Þ5 1 si =LðtÞ: ] 2
Proof of Lemma 4. We note that every integer d4st is in DðtÞ and by the choice of pkðtÞ ; every prime popkðtÞ is in DðtÞ: Thus for deDðtÞ and d4LðtÞ; d must reside in one of the following sets: D1 ¼ fd4LðtÞ: there exists m2 such that st om2 oLðtÞ and m2 j dg; D2 ¼ fd4LðtÞ: there exists p such that pkðtÞ opoLðtÞ and p j dg:
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
365
Then X X 1 1 X 1 X 1 4 4 þ : d d4LðtÞ;deDðtÞ d d2D d d2D d d2D 2
We first estimate X 1 4 d d2D 1
1
2
P
1 d2D1 d :
X st om2 4LðtÞ
1 m2
2 ½LðtÞ=m X
j¼1
1 j
4 ðlog LðtÞ log sðtÞ þ 1Þ
X st om2 oLðtÞ
4 ðlog LðtÞ log sðtÞ þ 1Þ
X
1 m2
pffiffiffiffiffiffi pffiffiffi st omo LðtÞ
1 1 m1 m
! log pkðtÞ log pkðtÞ ¼O ¼O : pffiffiffiffi st ðpkðtÞ Þ1=4 P We next estimate d2D2 d1: To do this, we break the interval ½pkðtÞ ; LðtÞ into three subintervals of the form ! " ! " LðtÞ LðtÞ LðtÞ LðtÞ ; LðtÞ : pkðtÞ ; ; ; ; log LðtÞ ðlog LðtÞÞ2 ðlog LðtÞÞ2 log LðtÞ Let 1
QðmÞ ¼ LðtÞðlog LðtÞÞm ; RðmÞ ¼ LðtÞðlog LðtÞÞ
mþ1 m ;
PðmÞ ¼ fp: QðmÞopoQðm þ 1Þg; RðmÞ ¼ fp: RðmÞopoRðm þ 1Þg: Then X 1 ¼ d d2D 2
¼
½LðtÞ=p 1 X 1 p j¼1 j opoLðtÞ
X
pkðtÞ
X pkðtÞ opoR1
¼ S1 þ S2 ;
0 1 ½LðtÞ=p 1 X 1 X 1 ½LðtÞ=p X 1 X 1 ½LðtÞ=p 1 X 1 X @ A þ þ p j¼1 j m¼1 p2RðmÞ p j¼1 j p2PðmÞ p j¼1 j say:
366
HISASHI YOKOTA
We shall estimate S1 and S2 : Note that by [7], Oð1=log2 xÞ: Then we have S1 ¼
X pkðtÞ opoR1
P
1 p4x p
¼ log2 x þ B þ
½LðtÞ=p X 1 X 1 1 4ðlog LðtÞ log pkðtÞ þ 1Þ p j¼1 j p pkðtÞ opoR1 t
1 ¼ ðlog LðtÞ log pkðtÞ þ 1Þ log2 Rð1Þ log2 pkðtÞ þ O 2 log pkðtÞ
!!
Since LðtÞ ¼ pkðtÞ ðlog pkðtÞ Þ2 ðlog2 pkðtÞ Þ2 ; we have LðtÞ Rð1Þ ¼ 4pkðtÞ ðlog2 pkðtÞ Þ2 : 2 ðlog LðtÞÞ Thus,
! log3 pkðtÞ ðlog2 pkðtÞ Þ2 S1 ¼ ð2 þ oð1ÞÞ log2 pkðtÞ O ¼o : log pkðtÞ log pkðtÞ
For the estimate of S2 ; we note that 1 X 1 ðm þ 1Þ log2 LðtÞ k log2 RðmÞ ¼ log2 LðtÞ ; k m log LðtÞ k¼1 log2 QðmÞ ¼ log2 LðtÞ
1 X 1 log2 LðtÞ k : k m log LðtÞ k¼1
Then, X 1 1 ¼ log2 Rðm þ 1Þ log2 RðmÞ þ O 2 p log RðmÞ p2RðmÞ
!
! 1 X 1 mþ1 k mþ2 k log2 LðtÞ k ¼ k m mþ1 log LðtÞ k¼1 1 þ O 2 log RðmÞ
!
! 1 log2 LðtÞ 1 log2 LðtÞ 2 þO ¼ mðm þ 1Þ log LðtÞ m2 ðm þ 1Þ log LðtÞ
1 þO 2 log RðmÞ
!
:
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
and X 1 1 ¼ log2 Qðm þ 1Þ log2 QðmÞ þ O 2 p log QðmÞ p2PðmÞ
367
!
! k k ! 1 X 1 1 1 log2 LðtÞ k 1 ¼ þO k m mþ1 log LðtÞ log2 QðmÞ k¼1 ! 1 1 log2 LðtÞ 1 log2 LðtÞ 2 þO ¼ m m þ 1 log LðtÞ m2 ðm þ 1Þ log LðtÞ
! 1 þ O : log2 QðmÞ Thus, X 1 X 1 X 1 ½LðtÞ=p 4 p j¼1 j p2RðmÞ p p2RðmÞ
mþ1=m ½ðlog LðtÞÞ
X
j¼1
mþ1 log2 LðtÞ þ 1 4 m
¼
1 j
X 1 p p2RðmÞ
1 ðlog2 LðtÞÞ2 log2 LðtÞ þ O m2 log LðtÞ mðm þ 1Þ log LðtÞ log2 LðtÞ þO log2 RðmÞ
!
and 1=m
LðtÞÞ X 1 ½LðtÞ=p X 1 X 1 ½ðlogX 1 4 p j¼1 j p2PðmÞ p j j¼1 p2PðmÞ X 1 1 log LðtÞ þ 1 4 m 2 p p2PðmÞ 1 1 ðlog2 LðtÞÞ2 ¼ m2 mðm þ 1Þ log LðtÞ
þ O
! log2 LðtÞ log2 LðtÞ þO : mðm þ 1Þ log LðtÞ log2 QðmÞ
368
HISASHI YOKOTA
Now we evaluate S2 : 0 1 1 X 1 X 1 ½LðtÞ=p X 1 X 1 ½LðtÞ=p X @ A þ S2 ¼ p j p j j¼1 j¼1 m¼1 p2RðmÞ p2PðmÞ 1 X 2 1 ðlog2 LðtÞÞ2 4 m2 mðm þ 1Þ log LðtÞ m¼1 !! log2 LðtÞ log2 LðtÞ þ O þO : mðm þ 1Þlog LðtÞ log2 RðmÞ
We note that 1 X 2 1 2p2 1; ¼ 2 m mðm þ 1Þ 6 m¼1 ! 1 1 1 dx ¼ O ¼O : 2 log Rð1Þ log LðtÞ log Rð1Þ x
Z
1 X
1 ¼O 2 m¼1 log RðmÞ
1
Thus, 2 2 ðlog2 pkðtÞ Þ2 2p 6 ðlog2 LðtÞÞ2 p 3 þ oð1Þ þ oð1Þ S2 4 ¼ : 6 3 log LðtÞ logpkðtÞ This proves that for t large, X 1 p2 3 ðlog2 pkðtÞ Þ2 4 þ oð1Þ : d 3 log pkðtÞ d4LðtÞ deDðtÞ
On the other hand, X 1 X 1 5 5S2 d d2D d d4LðtÞ 2
deDðtÞ
and S2 ¼
1 X m¼1
0
1 X 1 X 1 ½LðtÞ=p X 1 X 1 ½LðtÞ=p @ A: þ p j¼1 j p2PðmÞ p j¼1 j p2RðmÞ
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
369
Since X 1 X 1 X 1 ½LðtÞ=p 5 p j¼1 j p2RðmÞ p p2RðmÞ
mþ2=ðmþ1Þ ½ðlog LðtÞÞ
X
j¼1
1 j
X mþ2 1 1 log2 LðtÞ þ g þ O ¼ mþ1 log LðtÞ p p2RðmÞ ðlog2 LðtÞÞ2 log2 LðtÞ ¼ þO mðm þ 1Þ log LðtÞ mðm þ 1Þ2 log LðtÞ mþ2
1 þ O 2 log RðmÞ
!
and X 1 ½LðtÞ=p X 1 X 1 5 p j¼1 j p2PðmÞ p p2PðmÞ
1=ðmþ1Þ ½ðlog LðtÞÞ
X
j¼1
1 j
¼
X 1 1 1 log2 LðtÞ þ g þ O mþ1 log LðtÞ p p2PðmÞ
¼
ðlog2 LðtÞÞ2 log2 LðtÞ þ O mðm þ 1Þ log LðtÞ mðm þ 1Þ2 log LðtÞ 1
! 1 þO ; log2 QðmÞ we have S2 5
1 X m¼1
¼
1 X m¼1
! ! ðlog2 LðtÞÞ2 log2 LðtÞ þO þ mðm þ 1Þ log LðtÞ log LðtÞ mðm þ 1Þ2 mðm þ 1Þ2 mþ2
1
m þ 3 ðlog2 LðtÞÞ2 mðm þ 1Þ2 log LðtÞ
! þO
log2 LðtÞ : log LðtÞ
We note that mþ3 mðm þ 1Þ
2
¼
3ðm þ 1Þ 2m mðm þ 1Þ
2
¼
3 2 : mðm þ 1Þ ðm þ 1Þ2
370
HISASHI YOKOTA
Thus,
S2 5
1 X m¼1
! 3 2 ðlog2 LðtÞÞ2 log2 LðtÞ þO mðm þ 1Þ ðm þ 1Þ2 log LðtÞ log LðtÞ
2 p ðlog2 LðtÞÞ2 ¼ 32 1 þ oð1Þ 6 log LðtÞ ¼
ðlog2 pkðtÞ Þ2 15 p2 þ oð1Þ : 3 log pkðtÞ
]
Proof of Lemma 5. We first note that for st prime fs1 ; s2 ; . . . ; st ; puðtÞ ; . . . ; pkðtÞ g ¼ fs1 ; s2 ; . . . ; st1 ; puðt1Þ ; . . . ; pkðt1Þ g [ fpkðt1Þþ1 ; . . . ; pkðtÞ g i
and for st ¼ p2 for some i51; fs1 ; s2 ; . . . ; st ; puðtÞ ; . . . ; pkðtÞ g ¼ fs1 ; s2 ; . . . ; st1 ; puðt1Þ ; . . . ; pkðt1Þ g [ fst ; pkðt1Þþ1 ; . . . ; pkðtÞ g:
Thus in either case, we have X 1 X 1 X 1 X ¼ d d4Lðt1Þ d d4LðtÞ d d4LðtÞ d4LðtÞ d2DðtÞ
d2Dðt1Þ
d2DðtÞ
þ
d2Dðt1Þ
X d4LðtÞ d2Dðt1Þ
4
1 d
X 1 1 d d4Lðt1Þ d d2Dðt1Þ
½LðtÞ=s ½LðtÞ=p LðtÞ X X 1 Xt 1 1 X 1 1 þ þ st j¼1 j pkðt1Þ op4pkðtÞ p j¼1 j j¼Lðt1Þ j
¼ S1 þ S2 þ S3 ;
say:
NUMBER OF INTEGERS REPRESENTABLE AS UNIT FRACTIONS
371
Now we estimate S1 ; S2 ; S3 : Since by Lemma 1, log pkðt1Þ 5log pkðtÞ 1 O 1=5 ; we have pkðtÞ ! ½LðtÞ=s log pkðtÞ 1 Xt 1 1 4 ðlog LðtÞ log st þ 1Þ ¼ O pffiffiffiffiffiffiffiffiffi ; S1 ¼ st j¼1 j st pkðtÞ X
S2 ¼
pkðt1Þ op4pkðtÞ
½LðtÞ=p X 1 X 1 1 4ðlog LðtÞ log pkðt1Þ þ 1Þ p j¼1 j p pkðt1Þ op4pkðtÞ
¼ ð2 þ oð1ÞÞ log2 pkðtÞ
2 log2 pkðtÞ log2 pkðt1Þ þ 2 log pkðt1Þ
!
log2 pkðtÞ ¼ ð4 þ oð1ÞÞ 2 ; log pkðtÞ S3 ¼
LðtÞ X j¼Lðt1Þ
1 1 ¼ log LðtÞ log Lðt 1Þ þ O j Lðt 1Þ
0
1 1 ¼ O@ 1=5 A: pkðtÞ Thus, we have X 1 X 1 log2 pkðtÞ 4ð4 þ oð1ÞÞ 2 : d d log pkðtÞ d4LðtÞ d4Lðt1Þ d2DðtÞ
d2Dðt1Þ
This completes the proof of Lemma 5.
]
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